CBSE 2025, Set 1

Question

The electric field at a point in a region is given by $\vec{E} = \alpha\dfrac{\vec{r}}{|\vec{r}|^3}$, where $\alpha$ is a constant and $r$ is the distance of the point from the origin. The magnitude of potential of the point is : A). $\dfrac{\alpha}{r}$ B). $\dfrac{\alpha r^2}{2}$ C). $\dfrac{\alpha}{2r^2}$ D). $-\dfrac{\alpha}{r}$

Edvaya · Accepted Answer

The correct answer is $\dfrac{\alpha}{r}$. The electric field magnitude is: $$E = \frac{\alpha}{r^2}$$ The potential is related to the electric field by: $$V = -\int_{\infty}^{r} E \, dr = -\int_{\infty}^{r} \frac{\alpha}{r^2} \, dr$$ $$V = -\alpha \left[-\frac{1}{r}\right]_{\infty}^{r} = -\alpha\left(-\frac{1}{r} + 0\right) = \frac{\alpha}{r}$$ The **magnitude** of the potential is $\dfrac{\alpha}{r}$. > **Note:** This electric field has the same form as that of a point charge ($\alpha = q/4\pi\varepsilon_0$). The potential $V = \alpha/r$ is positive and decreases with distance — consistent with a positive point charge.

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Physics > Electrostatics And Current Electricity > Electric Charges And Fields

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